Lagrangian for a charged particle in a magnetic field

[mmcf]

Hey I have a question that was almost completely answered in this thread:
https://www.physicsforums.com/showthread.php?t=116098

Specifically someone interpreted his question to be how exactly do you get that
the term in the Lagrangian corresponding to the magnetic potential is

-q\vec{v}\cdot\vec{A}

The only proofs I have seen rely on the Lagrangian principle. E.g.

The Lagrangian

L=T-V=\frac{1}{2}mu^2 + q\vec{u} \cdot \vec{A}

Subsituted in the Euler Lagrange equations

\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}

Using generalised coordinates x, y, z with generalized velocities \dot{x}, \dot{y}, \dot{z} and u^2=\dot{x}^2+\dot{y}^2+\dot{z}^2

yields

\frac{d}{dt}(m\dot{x}+qA_x)=q\frac{\partial}{\partial x}(\dot{x}A_x+\dot{y}A_y+\dot{z}A_z)
m \ddot{x}+q \frac{\partial A_x}{\partial t} =q(\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x})

or after rearranging

m \ddot{x}=-q(\frac{\partial \phi}{\partial x}+\frac{\partial A_x}{\partial t})+q[\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})]=qE_x+q(\dot{y}B_z-\dot{z}B_y)

Which is exactly the x-component of the usual Lorentz force. So the potential term V=-q\vec{u} \cdot \vec{A} can be motivated by the fact that it yields the right equation of motion.

Sadly, I had already worked up to the second to last line prior to his post, and I'm not clear
on what happened at that point. I would assume he added in \partial \phi/\partial x because the electric potential is zero and that expression will look more like the typical E_x that you see. The part that loses me though is how did he turn the term

\dot{x}\frac{\partial A_x}{\partial x}+\dot{y}\frac{\partial A_y}{\partial x}+\dot{z}\frac{\partial A_z}{\partial x} into \dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})

Just from the looks of things it might be that:
\dot{x}\frac{\partial A_x}{\partial x} + \dot{y}\frac{\partial A_x}{\partial y} + \dot{z}\frac{\partial A_x}{\partial z} = 0

I don't really see why that should be true, so is there another way he could have done that, or what is the reasoning behind that equation? Also, sorry in advance. I did take a look at Jackson but he jumps into some relativistic stuff that I'm not ready for right now and we don't have a copy of Franklin in our library.

 

[Count Iblis]

It's explained in the book: "The Classical Theory of Fields" by Landau and Lifschitz.

 

[Meir Achuz]

<br /> m\frac{d{\vec v}}{dt}=<br /> q[-\nabla\phi-\partial_t{\vec A}+{\vec v}\times(\nabla\times{\vec A})] <br /> = q[-\nabla\phi-\partial_t{\vec A}-({\vec v}\dot\nabla){\vec A}<br /> +\nabla({\vec v}\cdot{\vec A})]<br /> =\nabla[-q\phi+q{\vec v}\cdot{\vec A}]-q\frac{d{\vec A}}{dt}.<br />

 

[mmcf]

It's explained in the book: "The Classical Theory of Fields" by Landau and Lifschitz.

Alright thanks. It looks like my problem happened here:

\frac{d}{dt}(m\dot{x}+qA_x)=q\frac{\partial}{\partial x}(\dot{x}A_x+\dot{y}A_y+\dot{z}A_z)

when I assumed that you could take that derivative of A_x with respect to time partially. If I had taken a total derivative I'd have seen instead:

m\ddot{x}+q\frac{\partial A_x}{\partial t} + q(\dot{x}\partial_x A_x + \dot{y}\partial_y A_x + \dot{z} \partial_z A_x) =q(\dot{x}\partial_x A_x +\dot{y}\partial_x A_y+\dot{z}\partial_z A_z)

and that gives me what I was hoping for:

m \ddot{x}=-q\frac{\partial A_x}{\partial t}+q[\dot{y}(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})+\dot{z}(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})]=qE_x+q(\dot{y}B_z-\dot{z}B_y)